Question

any one good at graphing quadratic functions if so respond and ill post the problems.

Answer 1

what do you need to know...?

Answer 2

how to graph these:
1. f(x)=-2x
2. f)x)=x^2-4x+4
3. f(x)=x^2-6x+8

Answer 3

ok... 1. is a straight line

f(x) = -2x use a table of values x: 1, 2, 3
y: -2 -4 -6

so the ordered pairs (1, -2), (2, -4) and (3, -6) could be plotted and a line drawn through them .

2. Can you factorise the quadratic...?

Answer 4

yes

Answer 5

what is the factorised form?

Answer 6

(x-2)(x-2)

Answer 7

yep

\[y = (x - 2)^2\]

so the graph has a root( or contacts the x-axis) when

\[(x - 2)^2 = 0\]

so a root exists at x = 2

the curve has a y-intercept when x = 0 so the y- intercept is y = 4

and the parabola is concave up. The general form is

\[y=ax^2 + bx + c\]

in your question a > 0 so its concave up...



the curve should touch the x-axis at x = 2